Question

How do you integrate `((x^2) / (16-x^3)^2) dx`?

Answer 1

`1/(3(16-x^3))+C`

Explanation:

Choose `t=16-x^3`, then
`dt=-3x^2dx => x^2dx=-dt/3`
`intx^2/(16-x^3)^2dx=int1/t^2(-dt/3)=-1/3intt^(-2)dt=`
`=-1/3 t^(-1)/-1+C=1/3 1/t+C=1/(3(16-x^3))+C`